Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 331 / 2

Linearization about $w=0$ removes the quadratic term and givesThe three homogeneous boundary conditions select the vertical normal modesTake a horizontal Laplacian eigenfunction satisfyingOn $W_{j}f$, the full Laplacian has eigenvalue $-(a^2+j^2{\pi }^2)$. The growth rate is consequentlyNeutrality occurs atFor fixed nonzero $a$, this increases strictly with $j$, so the first unstable vertical mode is $j=1$ and

At criticality choose the fundamental modeIts quadratic self-interaction satisfiesThe critical linear operator has eigenvalues $-64{\pi }^6$ on $\sin (2\pi z)$ and $-60(a^2+{\pi }^2{)}^3$ on $\cos (2ax)\sin (2\pi z)$. The slaved second-order correction is thereforewhere

Introduce a slow time $\tau$ and project the next-order equation onto the fundamental mode. Detuning $R-R_c$ contributes $a^2(R-R_c)A$, while the interaction of $w_1$ with $w_2$ has fundamental componentThe solvability condition is the Landau amplitude equation

SetThe amplitude equation is $\dot{A}=\mu A-g{A}^3$. Its equilibria arefor every $R$, and, when $R>R_c$,For $R<R_c$, the zero branch is stable and every sufficiently small amplitude decays to zero. At $R=R_c$ it loses stability. For $R>R_c$, zero is unstable and the two nonzero branches are stable; positive initial amplitudes approach $A_{+}$ and negative ones approach $A_{-}$. The bifurcation diagram is therefore a supercritical pitchfork.